Consider the following problem:
$n$ people participate in a project. Each of them can invest any sum of money in it. If the total number of money invested exceeds $c$ the project becomes completed and each participant gains $m$ revenue no matter how much they actually invested. If it does not, the project is not completed and everyone lose their investment without receiving anything. None of the participants knows how much anyone else has invested. Will the project be completed?
Here we are looking for symmetric Nash Equilibria. The strategy of a player is $x$ - the amount of money to invest, and the revenue function is:
$$\begin{cases} m - x & \quad nx \geq c \\ -x & \quad nx < c \end{cases}$$
Note, that it has only two local maxima: $x = 0$ and $x = \frac{c}{n}$.
From that we can see that those projects can be divided into three classes: «useless», for which $m < \frac{c}{n}$, «maybe useful», for which $\frac{c}{n} \leq m \leq c$, and «definitely useful», for which $c < m$.
If the project is «useless», the answer is no. Paying nothing is the only symmetric Nash Equilibrium here (because completion of the project will definitely lead by pigeonhole principle to someone wasting money)
If the project is «maybe useful», the answer is maybe. Now, there are two symmetric Nash equilibria. One is when everyone pays exactly $\frac{c}{n}$ and the project is completed (if someone pays less the project will no be completed and there will be no profit from it, if someone pays more, they waste the excess money they paid). The other is when no-one pays anything (as none can cover the cost of the project by themselves without losing money).
If the project is «definitely useful» then the answer is yes. Everyone paying $\frac{c}{n}$ is still a Nash equilibrium, but no-one paying anything isn’t (because each participant can cover all costs of the project by themselves and still profit from it).
Now, let’s make the problem a bit more difficult:
$n$ people participate in a project. Each of them can invest any sum of money in it. If the total number of money invested exceeds $c $ the project becomes completed and for all $i \leq n$ $i$-th participant gains $x_i $ revenue no matter how much they actually invested. If it does not, the project is not completed and everyone lose their investment without receiving anything. Participants know only their value of $x_i$ and assume the values of everyone else to be independent and uniformly distributed on $[0; 1]$. None of the participants knows how much anyone else has invested. Will the project be completed?
Now we are looking for a symmetric Bayes-Nash equilibrium. Here, the strategy of each participant will be the function $h_i: \mathbb{R}_+ \to \mathbb{R}_+$ of the amount of money to invest, depending on $x_i$.
Even the case when $n = 2$ is very complex:
Suppose $x_1 = t$ and they chose to invest $s$. Suppose, the second player follows strategy $h$. Then the expected revenue of the first player is:
$tP(h(x_2) \geq c - s) - s = t (1 - h^{-1}(c - s)) - s$
Let's find the $s$ for which it is maximal.
$\frac{\delta}{\delta s}(t (1 - h^{-1}(c - s)) - s) = \frac{1}{h'(h^{-1}(c - s))} - 1$
That means, maximum is reached when $h'(h^{-1}(c - s)) = t$. The symmetric Nash equilibrium is reached in case $s = h(t)$. So we have:
$h'(h^{-1}(c - h(t))) = t$
And now the problem is reduced to a differential equation I have no idea how to solve. Neither do I have any ideas how to generalise this method for $n \geq 3$.